How does an LED turn electrical energy into light?

An LED, or light-emitting diode, produces light in a semiconductor rather than by making a filament extremely hot. Different regions of the semiconductor provide mobile electrons and holes. A hole is an unoccupied electronic state that behaves as a positive charge carrier; it is not a tiny physical hole drilled in the material.

A forward voltage injects carriers toward an active region. When an electron recombines with a hole, the electronic system loses energy. In a radiative event, that energy leaves as a photon. Materials and device structures are selected to make this light-producing process effective. An ordinary silicon diode does not become an efficient lamp simply because it carries current.

The emitted photon energy is related to the semiconductor’s energy gap. A useful ideal estimate is wavelength in nanometers ≈ 1240 divided by photon energy in electronvolts. A 2.0-electronvolt photon corresponds to about 620 nanometers, while a 3.0-electronvolt photon corresponds to about 413 nanometers. A larger photon energy means a shorter wavelength. These are illustrative monochromatic estimates; real LEDs emit a band of wavelengths.

Not every recombination emits useful escaping light. Energy can be transferred to the material without a photon, and some generated light is trapped or absorbed. Practical lamps therefore also need electrical control and thermal management. Many white lamps combine a blue LED with a phosphor that converts some light to longer wavelengths. Their white appearance comes from the combined spectrum, not from a white filter or the semiconductor emitting every visible wavelength equally.